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4 years ago

80 POINTS!!! ALL PLATO USERS: asap!!!

1 answer:

Crank4 years ago

4 0

This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and

rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course

prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.

Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience

expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness

in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of

academic and professional writing; they will learn to read complex texts with understanding and to write prose of

sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,

but not exclusively, from American writers. Students who enroll in the class will take the AP examination.

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Determined from the sum of potential and kinetic energy in a substance

sattari [20]

Answer:

heat.

hope i helps

7 0

3 years ago

Why pie bond are not perticipate in hybridaization

DedPeter [7]

The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.

3 0

3 years ago

Asap please

MissTica

Answer:

When $\Delta H^o < 0$ and $\Delta S^o > 0$

Explanation:

The spontaneity of a reaction can be determined by the sign of the Gibbs free energy change. For a spontaneous reaction, the change in Gibbs free energy should be negative, meaning:

$\Delta G^o = \Delta H^o - T\Delta S^o < 0$

Since the temperature value here corresponds to the absolute temperature, this implies that T > 0 for any T. Therefore, to have a negative difference for any temperature value, the first term, the change in enthalpy, should be negative and the change in entropy should be positive, so that we always subtract a positive number from a negative number. This corresponds to a negative value in $\Delta G^o$ .

6 0

4 years ago

If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em

erica [24]

Answer: Wavelength associated with the fifth line is 397 nm

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $10973731.6m^{-1}$

$n_f$ = Higher energy level = 7

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

$\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.52\times 10^{6}m$

$\lambda}=3.97\times 10^{-7}m=397 nm$ $1nm=10^{-9}m$

Thus wavelength λ associated with the fifth line is 397 nm

7 0

4 years ago

Calculate the percent yield of benzyl alcohol Sue obtained when doing the following reaction. Sue started with 2.1 g of benzoic

Pachacha [2.7K]

Answer:

The percent yield is 116.6 %

Explanation:

Step 1: Data given

Mass of benzoic anhydride = 2.1 grams

Mass of sodium borohydride = 0.15 grams

Molar mass of benzoic anhydride = 226.23 g/mol

Molar mass of NaBH4 = 37.83 g/mol

Mass of benzyl alcohol produced = 0.5 grams

Step 2: The balanced equation

C14H10O3 + NaBH4 → C7H6O2 + C7H8O + NaB

Step 3: Calculate moles of C14H10O3

Moles C14H10O3 = Mass / molar mass

Moles C14H10O3 = 2.10 grams / 226.23 g/mol

Moles C14H10O3 = 0.00928 moles

Step 4: Calculate moles NaBH4

Moles NaBH4 = 0.150 grams / 37.83 g/mol

Moles NaBH4 = 0.00397 moles

Step 5: Calculate limiting reactant

For 1 mol of C14H10O3 we need 1 mol of NaBH4 to produce 1 mol of C7H8O

NaBH4 is the limiting reactant. It will completely be consumed ( 0.00397 moles).

C14H10O3 is in excess. There will react 0.00397 moles.

There will remain 0.00928 - 0.00397 = 0.00531 moles

Step 6: Calculate moles of C7H8O

For 1 mol of C14H10O3 we need 1 mol of NaBH4 to produce 1 mol of C7H8O

For 0.00397 of C14H10O3 we need 0.00397 mol of NaBH4 to produce 0.00397 mol of C7H8O

Step 7: Calculate mass of C7H8O

Mass of C7H8O = Moles * molar mass

Mass of C7H8O = 0.00397 moles * 108.14 g/mol

Mass of C7H8O = 0.429 grams = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield/ theoretical yield) * 100%

% yield = 0.5/0.429) *100 %

% yield = 116.6 %

The percent yield is 116.6 %

3 0

4 years ago